Making graphs solvable in peg solitaire

In 2011, Beeler and Hoilman generalized the game of peg solitaire to arbitrary connected graphs. Since then peg solitaire has been considered on quite a few classes of graphs. Beeler and Gray introduced the natural idea of adding edges to make an unsolvable graph solvable. Recently, the graph invariant ms( G ) , which is the minimal number of additional edges needed to make G solvable, has been introduced and investigated on banana trees by the authors. In this article, we determine ms( G ) for several families of unsolvable graphs. Furthermore, we provide some general results for this number of Hamiltonian graphs and graphs obtained via binary graph operations.


Introduction
In [3], Beeler and Hoilman introduced the game of peg solitaire on graphs as a generalization of the classical peg solitaire game: Given a connected, undirected graph G = (V, E), we can put pegs in the vertices of G. Given three vertices u, v, w with pegs in u and v and a hole in w such that uv, vw ∈ E, we can jump with the peg from u over v into w, removing the peg in v (see Figure 1). This jump will be denoted as u · v · w. In general, we begin with a starting state S ⊂ V of vertices that are empty (i.e., without pegs). A terminal state T ⊂ V is a set of vertices that have pegs at the end of the game such that no more jumps are possible. A terminal state T is associated to a starting state S, if T can be obtained from S by a series of jumps. We will always assume that the starting state S consists of a single vertex.
The goal of the original game is to remove all pegs but one. This is not possible for all graphs. Therefore, we use the following notation. A graph G is called • solvable, if there is some v ∈ V such that the starting state S = {v} has an associated terminal state consisting of a single vertex.
• freely solvable, if for all v ∈ V the starting state S = {v} has an associated terminal state consisting of a single vertex.
• k-solvable, if there is some v ∈ V such that the starting state S = {v} has an associated terminal state consisting of k vertices.
• strictly k-solvable, if G is k-solvable but not -solvable for any < k. In that case G has solitaire number Ps(G) = k.
In 2016 [1], Beeler and Gray considered the natural question of determining the minimum number of edges necessary to guarantee the solvability of a connected graph. Furthermore, they posed the question of how much the addition of edges can influence the solvability of a graph. In [8], the authors defined the smallest number ms(G) of edges that have to be added to a graph G to make it solvable and provided an example showing that the solvability might be improved arbitrarily good with the addition of just one edge. Since many unsolvable graphs exist, it seems natural trying to compute ms(G) for these graphs. We do this for several graph classes in Section 2 and provide general results in Section 3. First, we start with two rather obvious, but important facts.
Every complete graph is solvable except for K 1 , which cannot have a starting state and an associated terminal state both of size one. Therefore, ms(G) exists for every graph G = K 1 (and thus we exclude the case G = K 1 whenever considering ms(G)). Furthermore, we have the following relationship between ms(G) and Ps(G). Proof. The cases |V (G)| < 3 and Ps(G) = 1 are trivial, hence we assume |V | ≥ 3 and Ps(G) ≥ 2. Let T = {t 1 , t 2 , . . . , t Ps(G) } be a terminal state of G with minimal number of pegs. We find some u ∈ V \ T with ut 1 ∈ E or ut 2 ∈ E, w.l.o.g. assume ut 1 ∈ E. After adding the edge t 1 t 2 , we can jump t 2 · t 1 · u. Next we add ut 3 (unless it does already exist) and jump t 3 · u · t 1 . Continuing in this alternating manner (adding t 4 t 1 , t 5 u and so on) yields a terminal state with one peg after adding at most Ps(G) − 1 edges.

Graph classes
In this section we determine ms(G) for some graph classes. We start with trivial results where ms(G) is either 0 or 1. These follow from known results on Ps(G) for the respective graphs together with Proposition 1.1.
Proposition 2.1. Let K n be the complete graph on n vertices, K m,n be the complete bipartite graph on m + n vertices, P n the path on n vertices, C n the cycle on n vertices and W (B) the windwill with B blades. Table 1 gives ms(G) for these graphs. To determine ms(G) for windmills with pendants and for stars, we define a more general version of generalized windmills. Let P, B ∈ Z ≥0 . A general windmill W * (P, B) with B blade vertices and P pendant vertices is a graph G with a vertex u that is adjacent to exactly P pendant vertices and B vertices that lie in blades, i.e., the induced subgraph defined by these vertices is a disjoint union of paths. Figure 2 shows a windmill W * (3, 7). Note that the parameters B and P do not fully characterize a general windmill, since, for example, the windmill W * (1, 4) could have 4 vertices lying in one blade of length 4 or two times two vertices lying in blades of length 2. This will not be a problem for us since we are only interested in the total number of blade vertices. Note also that in [4] the windmill W (P, B) has 2B blade vertices (which by definition induce a union of B paths of length 2), whereas W * (P, B) has only B blade vertices (this is due to the fact that we do not make assumptions on the size of the blades).
We begin with a result about solvability of general windmills. We assume (P, B) / ∈ {(1, 0), (2, 0)} since otherwise we are dealing with the paths P 2 resp. P 3 . To deal with general windmills and double stars we need the following lemma.
Proof. If P > B, Lemma 2.1 immediately yields the unsolvability of W * (P, B) since every pendant vertex is a leaf. If B ≥ P , we can use the same strategy as in [4] to solve the graph: Start with a hole in a pendant vertex. Jump from another pendant vertex over the centre into the hole. From now on, whenever the centre is empty, jump from two adjacent blade vertices into the centre. If the centre is not empty, jump from a pendant vertex over the centre into a blade vertex (in such a way that adjacent blade vertices are filled with pegs). Iterating this process yields the solvability.
Remark 2.1. Using the above strategy, we can also show that Ps(W * (P, This gives the following result about ms(G) for general windmills.
Proof. Note that if we add any non-existent edge to a graph W * (P, B), there are four possibilities: 1. We add an edge between two pendant vertices. This results in a graph W * (P − 2, B + 2). 2. We add an edge between a pendant vertex and a blade vertex. This yields a graph W * (P − 1, B + 1). 3. We add an edge between two blade vertices lying in disjoint blades. This gives us a graph W * (P, B). 4. We add an edge between two blade vertices lying in the same blade. The resulting graph will not be a general windmill any more but has the same number of leaves.
Due to Lemma 2.1, we need to add edges such that the number of leaves gets reduced. This means that the fourth option will never increase the solvability. Hence, we only need to consider the first three types of edges. Therefore, adding an edge will always yield another general windmill. Then, ms(W * (P, B)) is the least number of edges that result in a graph W * (P , B ) with B ≥ P . This implies that the best option is to join two pendant vertices with each edge added. Every such edge lowers Ps(W * (P, B)) by 4 (except for potentially the last edge). Thus, ms(W * (P, B)) is the smallest k such that P − B + 1 − 4k ≤ 1, i.e., ms(W * (P, B)) = P −B
Since K 1,n is a general windmill W * (n, 0) and W (P, B) is a general windmill W * (P, 2B), we get the following result. Corollary 2.1.
• Let K 1,n be the star graph with n leaves. If n ≥ 3, we have ms(K 1,n ) = n 4 .
We now turn our attention to double stars. The double star DS(L, R) is the union of the stars K 1,L and K 1,R together with an edge connecting the centres of the two stars. The following result on their solvability is known.  This idea does not work for R − L ≡ 0, 1 mod 4 (let d denote this remainder) since the last equation does not hold in those cases. We modify the proof slightly and use the fact ms(DS(1, 5)) = ms(DS(1, 6)) = 1. In both cases we start with solving a copy of DS(L − 1, L − 1), leaving pegs in one left pendant vertex, say 1 , in R − L + 1 right pendant vertices, denoted by r 1 , r 2 , . . . , r R−L+1 , and in the right centre c R . We jump r 1 · c R ·c L . Using the idea from Theorem 2.2 in [4] (see the proof of Proposition 2.2), we can, after adding R−L−d 4 edges r R−L+1 r R−L , r R−L−1 r R−L−2 , . . . , reduce the (solvable) windmill subgraph, which is induced by the vertices c R and r i for i ∈ [2 + d, R − L + 1], such that only pegs in r 2+d , r 3+d , r R−L+1 , r R−L remain. The subgraph induced by 1 , c L , c R and r i for i ∈ [2, 3 + d] ∪ {R − L + 1, R − L}, which contains the last remaining pegs and has a hole in c R , is solvable.

Note that ms(G) ≤ ms(H) holds if H is a spanning subgraph of G. On the other hand, if
Ps(G) ≤ Ps(G ), but no relationship between G and G is known, we cannot conclude anything about the relationship of ms(G) and ms(G ). For example, if B n,k denotes the banana tree on n stars, we have Ps(B 2,k ) = 2k − 2 and ms(B 2,k ) ≤ 2 (see [8]), but Ps(K 1,k ) = k − 1 and ms(K 1,k ) = k 4 . Furthermore, it is not true that the edges which have to be added to make G solvable have to connect vertices of a best possible terminal state. Therefore, in general, we have to start with adding edges instead of solving the original graph first and adding edges later (again the banana tree B 2,k is a nice example for this phenomenon). Now we turn to some general bounds for ms(G). Since every path is at least 2-solvable, the following result is immediate (see [3,Corollary 2.5] for some connections between solvability and the existence of a Hamiltonian cycle). This raises the question of how many edges have to be added to a graph to get a Hamiltonian path. There are various criteria (found in many books and web sources) for a graph G which guarantee that G has a Hamiltonian path. Together with Proposition 3.1 these may be used to give bounds on ms(G) for a given graph G. We will only show a connection to the path partition number of a graph (there are certainly many more). A path partition of a graph G is a set of paths such that every vertex of G belongs to exactly one path; the minimum cardinality, denoted by π p (G), of such a partition is called the path partition number of G [12]. Proof. Start with a minimal path partition P = {P (1) , P (2) , . . . , P (k) } of G. For each i = 1, 2, . . . , k, let P (i) = {p for i = 1, 2, . . . , k − 1 yields a Hamiltonian path which is solvable after adding at most one more edge.
To end the section, we will give some results about binary graph operations. Since the join G + H, which is G ∪ H together with additional edges connecting every pair of vertices g, h with g ∈ V (G) and h ∈ V (H), of any two graphs G and H with |V (G)|, |V (H)| ≥ 2 is solvable [3, Theorem 2.7], we have ms(G + H) = 0 in that case. The special case, where at least one of the graphs is K 1 , can be dealt with in the following way. As usual G[W ] denotes the subgraph of G induced by some set W ⊆ V (G).  Let G be a star, the union of two stars or the union of at least 3 non trivial stars. Then G + K 1 is solvable.
Proof. The special case G = 2K 1 yields G + K 1 = P 3 , which is solvable, hence from now on G is not of this form.
Let w denote the vertex corresponding to K 1 and V 1 , V 2 , . . . , V be the vertex sets of the stars from G such that and v 1,i is adjacent to all the vertices in V i . Start with a hole in w.
As long as V i and V j (i = j) with each of them containing at least 3 pegs exist, we carry out a double star purge on V i ∪ V j ∪ {w} (with centres w and v 1,i or v 1,j , depending on the size of V i and V j ) until there is a hole in w and at least one of V i and V j contains exactly two pegs (one of them being in the star centre v 1,i resp. v 1,j ). This results in a configuration, where we have a hole in w and all V i with i = j for some j ∈ {1, . . . , } contain exactly two pegs.
If V j contains exactly two pegs, we are done, since we find a solvable windmill subgraph. Otherwise, we can easily reduce the number of pegs in V j using w and induced cycles of length 3. Again, we obtain a solvable windmill subgraph.
holds. If s i = 2 for every i ∈ {1, . . . , }, G + K 1 contains a solvable windmill (where w, the vertex corresponding to K 1 in G + K 1 , is the centre and every V i forms a blade) because of (2).
Otherwise we can, starting with a hole in u 1 and jumping u 2 · w · u 1 , remove pegs from the sets V i with s i ≥ 3 and from leaves of G + K 1 using double star purges on the subgraphs induced by {u j , u j+1 , . . . , u j+s i −2 } ∪ {w} ∪ (V i \ {v 2,i }) (pick the smallest j such that u j contains a peg) until one of two configurations is reached.
• No V i with more than two pegs exists. Again, we find some solvable windmill subgraph (which is a subgraph of the graph induced by all vertices with pegs together with w).
• There is a hole in every u i and at least one V i contains more than 2 pegs. This can be solved by Lemma 3.2.
Combining this result and Lemma 2.1, we get the following proposition.
Proposition 3.4. Let G be a graph with k isolated vertices. Then For graphs G and H we denote the Cartesian product of G and H by G H and use the (common) notation (g, h) ∈ V (G H) for the vertex induced by g ∈ V (G) and h ∈ V (H).
Beeler and Hoilman showed that G H is solvable if G and H are solvable or if G is solvable and H is distance 2-solvable (meaning it is 2-solvable and the terminal vertices are at distance 2) or if both are 2-solvable [3]. The authors of this paper proved that P 2 G is solvable for any connected graph G and conjectured Ps(G H) = 1 for any two non trivial connected graphs G and H [10]. Hence, we suggest ms(G H) = 0 in that case. This statement seems out of reach at the moment. Since ms(G H) can be a lot smaller than ms(G) and ms(H) (for example if both are stars [10]), lower bounds seem difficult to achieve. Upper bounds are mostly trivial, hence we will not continue exploring Cartesian products in this article (although the above mentioned conjecture should definitely be investigated further).
If G ∪ H denotes the union of G and H, we have ms(G ∪ H) ≤ ms(G) + ms(H) + 2. To verify this, start with solving G using ms(G) additional edges. Let w ∈ H be any vertex such that H can be solved starting with a hole in w when adding ms(H) edges (and add these to G ∪ H).
Let v ∈ H be any neighbour of w. Connect the terminal vertex t of G with v by an additional edge and jump v · t · u for some u ∈ G. Next, add an edge between u and w and jump u · w · v. Now solve H.
Iterating this process gives the following result.

Open problems
Since the definition of ms(G) is new, there are some more questions that naturally arise. For instance, ms(G) could be determined for other classes of graphs (for example caterpillars, banana trees, or trees of diameter 4).
One might also define the number ms * (G) to be the minimal number of edges that have to be added to make a graph freely solvable (since K n is freely solvable, this number exists and is clearly greater than or equal to ms(G)). It would be interesting to examine this number for certain graph classes, get general results and see how this quantity relates to ms(G).
It would also be interesting to connect ms(G) to the edge-critical graphs defined in [1]. Moreover, adding edges may yield solvable graphs even if the original graphs are disconnected. This gives the possibility to study peg solitaire on graphs for which it was previously not possible. Obtaining more results on disconnected graphs than the ones in the previous section would be a desirable goal.